3 Determination of one-site anisotropy energy

Suppose that we describe the properties of a magnetic system with an anisotropic Heisenberg Hamiltonian as follows:

$$\mathcal{H}=-\frac{1}{2}\sum_{i,j}\mathbf{S_{i}}\mathcal{J}_{ij}\mathbf{S_{j}}- \sum_{i}\mathrm{d(\mathbf{S_{i}})}$$ (144)

where $\mathcal{J}_{ij}$ and $\mathrm{d(\mathbf{S_{i}})}$ are matrices $3 \times 3$. If we assume that the on-site anisotropy is uniaxial and its easy axis is parallel to Z axis, we can rewrite the Hamiltonian as:

$$\mathcal{H} = -\frac{1}{2}\sum_{i,j}\mathbf{S_{i}}\mathcal{J}_{ij}\mathbf{S_{j}}- \sum_{i}\mathbf{S_{i}}\mathrm{d_{i}}\mathbf{S_{i}}$$ (145)

$$\mathrm{d_{i}} = \left( \begin{array}{ccc} 0 &0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & d^{zz}_{i}\\ \end{array} \right)$$ (146)

In spherical coordinates we can write $\mathbf{S}_{i}$ as a function of the polar and azimuthal angles $\theta_{i}$ and $\varphi_{i}$, respectively.

$$\mathbf{S}_{i}=(\sin\theta_{i}\cos\varphi_{i}, \sin\theta_{i}\sin\varphi_{i},\cos\theta_{i})$$ (147)

and evaluate the second derivatives with respect to the polar and azimuthal angles.

$$\frac{\partial^{2}\mathbf{S}_{i}}{\partial\theta_{i}^{2}} = -(\sin\theta_{i}\cos\varphi_{i}, \sin\theta_{i}\sin\varphi_{i},\cos\theta_{i})=- \mathbf{S}_{i}$$ (148)

$$\frac{\partial^{2}\mathbf{S}_{i}}{\partial\varphi_{i}^{2}} = -(\sin\theta_{i}\cos\varphi_{i}, \sin\theta_{i}\sin\varphi_{i},0)$$ (149)

If the spin i belongs to the plane XY then Eq. (A.25) is reduced to :

$$\frac{\partial^{2}\mathbf{S}_{i}}{\partial\varphi_{i}^{2}}=- \mathbf{S}_{i}$$ (150)

Now we evaluate the second derivatives with respect to the polar and azimuthal angles of the energy:

$$\frac{\partial^{2}E\{\mathbf{S}_{i}\}}{\partial\theta_{i}^{2}} = -\frac{\partial^{2}}{\partial\theta_{i}^{2}}\Bigl( \mathbf{S_{i}}... ...S_{i}}\mathcal{J}_{ij}\frac{\partial^{2}\mathbf{S}_{j}}{\partial\theta_{i}^{2}}$$ (151)

$$\frac{\partial^{2}E\{\mathbf{S}_{i}\}}{\partial\varphi_{i}^{2}} = -\frac{\partial^{2}}{\partial\varphi_{i}^{2}}\Bigl( \mathbf{S_{i}... ..._{i}}\mathcal{J}_{ij}\frac{\partial^{2}\mathbf{S}_{j}}{\partial\varphi_{i}^{2}}$$ (152)

Due to the fact that the spin variables are independent, the equations above can be reduced to:

$$\frac{\partial^{2}E\{\mathbf{S}_{i}\}}{\partial\theta_{i}^{2}} = -\frac{\partial^{2}}{\partial\theta_{i}^{2}}\Bigl( \mathbf{S_{i}}... ...artial^{2}\mathbf{S}_{i}}{\partial\theta_{i}^{2}}\mathcal{J}_{ij}\mathbf{S_{j}}$$ (153)

$$\frac{\partial^{2}E\{\mathbf{S}_{i}\}}{\partial\varphi_{i}^{2}} = -\frac{\partial^{2}}{\partial\varphi_{i}^{2}}\Bigl( \mathbf{S_{i}... ...rtial^{2}\mathbf{S}_{i}}{\partial\varphi_{i}^{2}}\mathcal{J}_{ij}\mathbf{S_{j}}$$ (154)

The free energy of the system $\mathcal{F}$ is related with the second derivatives above:

$$\mathcal{F}(\theta_{i},\varphi_{i})=\frac{\partial^{2}E\{\mathbf{... ...{2}\mathbf{S}_{i}}{\partial\varphi_{i}^{2}}\Bigr)\mathcal{J}_{ij}\mathbf{S_{j}}$$ (155)

Taking into account the hypothesis that the on-site anisotropy is uniaxial Eq. (A.22), this expression can be reduced to:

$$\mathcal{F}(\theta_{i},\varphi_{i})=-\Bigl( \frac{\partial^{2}}{\... ...{2}\mathbf{S}_{i}}{\partial\varphi_{i}^{2}}\Bigr)\mathcal{J}_{ij}\mathbf{S_{j}}$$ (156)

Therefore we can calculate the on-site anisotropy constant $d^{zz}_{i}$ evaluating the free energy at $\theta_{i}=\pi/2$

$$\mathcal{F}(\theta_{i},\varphi_{i})\Bigr\vert _{\theta_{i}=\pi/2}=2d^{zz}_{i}\cos(2\theta)\Bigr\vert _{\theta_{i}=\pi/2}=-2d^{zz}_{i}$$ (157)

with

$$d^{zz}_{i}=-\frac{1}{2}\mathcal{F}(\theta_{i},\varphi_{i})\Bigr\vert _{\theta_{i}=\pi/2}$$ (158)